Major Product Formation in Diazotization and Azo Coupling Reaction

To determine the correct structure of the major product $W$, let us break down the reaction sequence step by step:

Step 1: Formation of Intermediate $V$ (Diazotization)

When aniline ($\text{C}_6\text{H}_5\text{NH}_2$) is treated with sodium nitrite ($\text{NaNO}_2$) and hydrochloric acid ($\text{HCl}$) at $0^\circ\text{C}$, it undergoes a diazotization reaction to form benzenediazonium chloride:

$\text{C}_6\text{H}_5\text{NH}_2 + \text{NaNO}_2 + 2\text{HCl} \xrightarrow{0^\circ\text{C}} \text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{NaCl} + 2\text{H}_2\text{O}$

Thus, intermediate $V$ is benzenediazonium chloride ($\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^-$).

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Step 2: Formation of Major Product $W$ (Azo Coupling Reaction)

When intermediate $V$ reacts with 2-naphthol ($\beta$-naphthol) in the presence of sodium hydroxide ($\text{NaOH}$), an electrophilic aromatic substitution (azo coupling) reaction occurs:

1. Role of $\text{NaOH}$: In alkaline medium, 2-naphthol is deprotonated to form the 2-naphthoxide ion, which significantly activates the aromatic ring towards electrophilic attack due to strong resonance donation ($+M$ effect) of the phenoxide oxygen: $\text{Ar-OH} + \text{OH}^- \rightleftharpoons \text{Ar-O}^- + \text{H}_2\text{O}$

2. Regioselectivity: The weak electrophile, benzenediazonium ion ($\text{C}_6\text{H}_5\text{N}_2^+$), preferentially attacks the 1-position ($\alpha$-position adjacent to the $-\text{OH}$ group) rather than the 3-position or positions on the other ring.

   • Electrophilic substitution at $\text{C}_1$ yields a carbocation intermediate (Wheland complex) in which the aromaticity of the adjacent benzene ring remains completely intact.
   • Attack at $\text{C}_3$ interrupts the benzene ring's aromatic resonance, making $\text{C}_1$ kinetically and thermodynamically favored.

3. Deprotonation from $\text{C}_1$ restores aromaticity in the substituted ring, forming 1-(phenylazo)-2-naphthol (a red-orange azo dye) as the major product $W$:

$\text{C}_6\text{H}_5\text{N}_2^+\text{Cl}^- + \text{C}_{10}\text{H}_7\text{O}^- \xrightarrow{\text{NaOH}} \text{C}_6\text{H}_5-\text{N}=\text{N}-\text{C}_{10}\text{H}_6(\text{OH}) + \text{Cl}^-$

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Conclusion:

The structure corresponding to 1-(phenylazo)-2-naphthol is given in Option (A).

Correct Option: A